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Question

The locus of the midpoints of chords of the circle $${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-2=0$$ which makes an angle $$120^{0}$$ at the center is


A
x2+y22x2y+1=0
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B
x2+y2xy+1=0
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C
x2+y22x2y1=0
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D
None of these
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Solution

The correct option is A $${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+1=0$$
Given equation of circle is $${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-2=0$$
Let mid point of the chord $$AB$$ is $$(h,k)$$
Its center is $$(1,1)$$ and radius $$=\sqrt { 1+1+2 } =2=OB$$.
In $$\triangle OPB, \angle OBP ={30}^{0}$$
$$\displaystyle \therefore \sin { { 30 }^{ 0 } } =\frac { OP }{ 2 } \Rightarrow OP=1$$
Since $$OP=1\Rightarrow { \left( h-1 \right)  }^{ 2 }+{ \left( k-1 \right)  }^{ 2 }=1$$
$$\Rightarrow { h }^{ 2 }+{ k }^{ 2 }-2h-2k+1=0$$ 
$$\therefore$$ Locus of the mid point of chord is
$${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+1=0$$

387780_33218_ans.PNG

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